Difference between revisions of "1959 AHSME Problems/Problem 44"

Latest revision as of 11:27, 22 July 2024

Problem

The roots of $x^2+bx+c=0$ are both real and greater than $1$ . Let $s=b+c+1$ . Then $s$ : $\textbf{(A)}\ \text{may be less than zero}\qquad\textbf{(B)}\ \text{may be equal to zero}\qquad$

$\textbf{(C)} \text{ must be greater than zero}\qquad\textbf{(D)}\ \text{must be less than zero}\qquad \textbf{(E)}\text{ must be between -1 and +1}$

Solution

Let the roots of the quadratic be $p$ and $q$ . Then, by Vieta's Formulas, $b=-(p+q)$ and $c=pq$ . By substituting these values of $b$ and $c$ into our expression for $s$ , we see that $s=pq-p-q+1$ . By SFFT, $s=(p-1)(q-1)$ . From the problem, we know that $p$ and $q$ are both greater than $1$ , so $(p-1)$ and $(q-1)$ are necessarily positive. Thus, $s$ , the product of these two positive terms, $\fbox{\textbf{(C) }must be greater than zero}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 43	Followed by Problem 45
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
-{{AHSME 50p box|year=1959|num-b=41|num-a=43}}
+{{AHSME 50p box|year=1959|num-b=43|num-a=45}}
 {{MAA Notice}}

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Difference between revisions of "1959 AHSME Problems/Problem 44"

Latest revision as of 11:27, 22 July 2024

Problem

Solution

See also